I am very exited to send the interesting problem on analysis.
Consider H is a non-empty subset of S. If H is bounded from above (below) then there exists a supremum (an infimum). Now, I am looking the constructive proof of this statement. Also, I am looking for, since S is an ordered field, how can we prove S is isomorphic to R?
Kindly discuss the proofs of the above interesting problem.
work done:
H is bounded from above, there exists some h such that h= max {q \in Z| ($q_0$, $q_1$,…) \in H}. Of course then what to do I don't know much.
reference:
Conway and Knuth
EDIT
'Prove that any totally ordered field with the lub property is unique up to isomorphism.'
Best Answer
If $S$ is a totally ordered field with the lub-property, you will not be able to prove constructively that any nonempty subset $H\subset S$ which is bounded from above has a least upper bound. This is a property you assume that $S$ has, not a property you prove that $S$ has.
It is, however, possible to prove that any nonempty subset $H\subset\bf R$ which is bounded from above has a least upper bound. The proof will depend on how you have defined the set of real numbers $\bf R$. Note that there are several definitions possible, all of which are equivalent. (The most common ones are by using Dedekind cuts of rational numbers, or by using equivalence classes of Cauchy sequences of rational numbers.)
If $S$ is a totally ordered field with the lub-property then there is a unique order-preserving field isomorphism $\phi : {\bf R}\to S$. This means that $\phi$ will have the following properties: $$\phi \hbox{ is 1-1 and onto}$$ $$\phi(x+y)=\phi(x)+\phi(y)$$ $$\phi(xy)=\phi(x)\phi(y)$$ $${\rm If\ }x<y{\rm\ then\ }\phi(x)<\phi(y)$$ In order to prove that there exists such a map $\phi$, you follow the steps given in the answer by Carl Mummert. There is some work involved here, if you want to write out all details, but it is healthy work which will deepen your understanding of all the involved concepts.
You begin by showing that if there is such a map $\phi$, then it is necessary that $\phi(0)=0_S$ and $\phi(1)=1_S$, where $0_S$ and $1_S$ are the zero element and the unit element of the field $S$, respectively. Hence we define $\phi(0)=0_S$ and $\phi(1)=1_S$.
Now use the field axioms to define $\phi(x)$ for any rational $x$. Show that this can only be done in one way if $\phi$ is to be a field isomorphism. Show that $\phi$ will then be 1-1 on $\bf Q$. Also show that $\phi$ will preserve the ordering on rational numbers, so that if $x,y\in\bf Q$ and $x<y$ then $\phi(x)<\phi(y)$.
You now need to extend $\phi$ so that it becomes a map which is defined on all of $\bf R$, and not only on $\bf Q$. I'm sure it can be done in several (equivalent) ways, but the following seems natural: If $x\in\bf R$, let $E=\{y\in {\bf Q}: y<x\}$. Note that $E$ is bounded above in $\bf R$, and that $x$ is a least upper bound of $E$. Show that $\phi(E)$ is bounded above in $S$, and define $\phi(x)$ to be the least upper bound of $\phi(E)$. Show that this does not change $\phi(x)$ in the case when $x\in\bf Q$. Show that the extension still has the properties $\phi(x+y)=\phi(x)+\phi(y)$, $\phi(xy)=\phi(x)\phi(y)$, and that it still is order-preserving and 1-1.
You now have an order-preserving injective map $\phi:{\bf R}\to S$ which preserves the field structure. The final step is to show that it is necessarily surjective. Let $s\in S$, we need to find $x\in \bf R$ such that $\phi(x)=s$. First show that there are $a,b\in\bf R$ such that $\phi(a)<s<\phi(b)$. Define $E=\{y\in {\bf R} : \phi(y)<s\}$. Show that $E$ is bounded above, and let $x$ be the least upper bound of $E$. Show that $\phi(x)=s$.